A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is

Question

A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is (Given: $R = 8.31$ $J$ $k^{- 1}$ $m o l^{- 1}$ ) [2024]

Smart Achievers · Accepted Answer

(2)

Given : $∆ Q = 48$ $J$

$1^{s t}$ law of thermodynamics, $∆ Q = ∆ U + ∆ W$

$\Rightarrow 48 = n C_{v} ∆ T + ∆ W$

$\Rightarrow 48 = (1) (\frac{3 R}{2}) (2) + ∆ W$

$\Rightarrow ∆ W = 48 - 3 \times R$

$\Rightarrow ∆ W = 48 - 3 \times (8.3)$

$\Rightarrow ∆ W = 23.1$ $J$

Solution

Q.
A total of 48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2°C. The work done by the gas is (Given: $R = 8.31$ $J$ $k^{- 1}$ $m o l^{- 1}$ ) [2024]

1 48 J

2 23.1 J

3 24.9 J

4 72.9 J

Ans.
(2)

Given : $∆ Q = 48$ $J$

$1^{s t}$ law of thermodynamics, $∆ Q = ∆ U + ∆ W$

$\Rightarrow 48 = n C_{v} ∆ T + ∆ W$

$\Rightarrow 48 = (1) (\frac{3 R}{2}) (2) + ∆ W$

$\Rightarrow ∆ W = 48 - 3 \times R$

$\Rightarrow ∆ W = 48 - 3 \times (8.3)$

$\Rightarrow ∆ W = 23.1$ $J$

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